package com.cg.offer;

import org.junit.Test;

import java.util.Stack;

/**
 * 剑指Offer 30.包含min函数的栈
 *
 * @program: LeetCode->Offer_30
 * @description: 剑指Offer 30.包含min函数的栈
 * @author: cg
 * @create: 2022-03-10 14:08
 **/
public class Offer_30 {
    @Test
    public void test30() {
        MinStack minStack = new MinStack();
        minStack.push(-2);
        minStack.push(0);
        minStack.push(-3);
        System.out.println(minStack.min());
        minStack.pop();
        System.out.println(minStack.top());
        System.out.println(minStack.min());
    }
}

/**
 * 定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
 * <p>
 * 示例:
 * MinStack minStack = new MinStack();
 * minStack.push(-2);
 * minStack.push(0);
 * minStack.push(-3);
 * minStack.min();   --> 返回 -3.
 * minStack.pop();
 * minStack.top();      --> 返回 0.
 * minStack.min();   --> 返回 -2.
 * <p>
 * 提示：
 * 各函数的调用总次数不超过 20000 次
 */
class MinStack {

    Stack<Integer> stack1, stack2;

    /**
     * initialize your data structure here.
     */
    public MinStack() {
        stack1 = new Stack<>();
        stack2 = new Stack<>();
    }

    public void push(int x) {
        stack1.push(x);
        if (stack2.isEmpty()) {
            stack2.push(x);
        } else if (x <= stack2.peek()) {
            stack2.push(x);
        }
    }

    public void pop() {
        if (stack1.pop().equals(stack2.peek())) {
            stack2.pop();
        }
    }

    public int top() {
        return stack1.peek();
    }

    public int min() {
        return stack2.peek();
    }
}